United States presidential election in New Jersey, 1836

Main article: United States presidential election, 1836

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The 1836 United States presidential election in New Jersey took place between November 3 and December 7, 1836, as part of the 1836 United States presidential election. Voters chose eight representatives, or electors to the Electoral College, who voted for President and Vice President.

New Jersey voted for the Democratic candidate, Martin Van Buren, over Whig candidate William Henry Harrison. Van Buren won New Jersey by a margin of 1.06%.

Results

United States presidential election in New Jersey, 1836^[1]
Party		Candidate	Running mate	Popular vote		Electoral vote
Party		Candidate	Running mate	Count	%	Count	%
	Democratic	Martin Van Buren of New York	Richard M. Johnson of Kentucky	26,137	50.53%	8	100.00%
	Whig	William Henry Harrison of Ohio	Francis Granger of New York	25,592	49.47%	0	0.00%
Total				51,729	100.00%	8	100.00%

References

↑ "1836 Presidential General Election Results - New Jersey". U.S. Election Atlas. Retrieved 23 December 2013.

State Results of the 1836 U.S. presidential election

Candidates	Martin Van Buren William Henry Harrison Hugh White Daniel Webster Willie Person Mangum

General articles	Election timeline

Local results	Alabama Arkansas Connecticut Delaware Georgia Illinois Indiana Kentucky Louisiana Maine Maryland Massachusetts Mississippi Missouri New Hampshire New Jersey New York North Carolina Ohio Pennsylvania Rhode Island South Carolina Tennessee Vermont Virginia

Other 1836 elections	House Senate Gubernatorial

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